\begin{enumerate}
\item 
The function $u: x \mapsto 1-x + \ln (2x)$ is defined on $]0,+\infty[$ and we obtain 
its derivative $u'$ :
\[u'(x)=-1+\frac{2}{2x}=-1+\frac{1}{x}=\frac{1-x}{x}. \]

\item 
Here we calculate the limits at 0 and $+\infty$ :

\item 
We have $x >0$ on $]0,+\infty[$, thus $u'(x)$ has the same sign as $1-x$.

The maximum of $u$ is $u(1)=\ln2$.

\end{enumerate}
